Output power measurement in an electrolytic capacitor-free LED driver solution

As a semiconductor device, the life of LED lamp beads is more than 50,000 hours. However, electrolytic capacitors are commonly used in LED lighting drive solutions, and their life is only 5,000~10,000 hours. There is a huge gap between the short life of electrolytic capacitors and the long life of LED lamp beads, which weakens the advantages of LED. Therefore, LED drive solutions without electrolytic capacitors are favored by the market.

Maxic Technology has launched an electrolytic capacitor-free LED driver solution based on MT7920 (see Figure 1). In this solution, after the full-bridge stack, a CBB high-voltage ceramic capacitor or a film capacitor with a smaller capacitance value is used to replace the high-voltage electrolytic capacitor, removing the electrolytic capacitor and also improving the power factor (PFC, which can be higher than 0.9 throughout the range of 85VAC~265VAC). The output capacitors C8 and C9 can be replaced by ceramic capacitors instead of electrolytic capacitors. This achieves a complete electrolytic capacitor-free solution.

Figure 1. Isolated LED driver solution based on MT7920.

When the output capacitors C8 and C9 use 470uF electrolytic capacitors and drive 6 LEDs, the measurement results are as follows:

• Input voltage Vin = 220VAC, input power Pin = 7.54W
• Output voltage Vo = 19.33V (multimeter reading)
• Output current Io = 327mA (multimeter reading)
• Output power Po = Vo * Io = 6.32W
• Efficiency η = 6.32/7.54 = 83.8%

The waveforms of the output voltage and current when using electrolytic capacitors are shown in Figure 2. It can be seen from the waveform that there are certain ripples in the output voltage and current. This is inevitable in the single-stage PFC constant current drive solution. Increasing the output capacitors C8 and C9 can further reduce the output ripple. At the same time, we noticed that the average values ​​of the current and voltage on the oscilloscope are basically the same as the readings of the multimeter. That is, the DC voltage and current values ​​measured by the multimeter are average values.

Figure 2. Current and voltage waveforms when the output uses electrolytic capacitors (470uF X 2)

(Ch1=blue: output voltage; Ch4=green: output current; math operation=red: Ch1*Ch4)

Furthermore, on the oscilloscope, the average value of 6.34W of the instantaneous power curve obtained by multiplying the output voltage by the output current is also substantially the same as the power calculated by multiplying the average voltage by the average current.

When the output capacitors C8 and C9 are 22?F ceramic capacitors and drive 6 LEDs, the measurement results are as follows:

• Input voltage Vin = 220VAC, input power Pin = 8.10W
• Output voltage Vo = 19.07V (multimeter reading)
• Output current Io = 334mA (multimeter reading)
• Output power Po = Vo * Io = 6.37W
• Efficiency η = 6.37/8.10 = 78.6%

The waveforms of the output voltage and current when using ceramic capacitors are shown in Figure 3. Compared with the case of using electrolytic capacitors, the input power increases by about 0.56W (8.10W – 7.54W), while the output power calculated by the multimeter reading remains basically unchanged (6.37W vs. 6.32W), resulting in a 5% decrease in efficiency. Is this really the case? Where did the 0.5W of power go?

Figure 3. Current and voltage waveforms when the output uses ceramic capacitors (22uF X 2).

(Ch1=blue: output voltage; Ch4=green: output current; math operation=red: Ch1*Ch4)

In Figure 3, the average value of the instantaneous power curve obtained by multiplying the output voltage and the output current is 6.86W, instead of 6.37W calculated by the average voltage and the average current. The difference between the two is 0.49W, which just makes up for the 0.56W added at the input. The new efficiency should be η = 6.37/8.10 = 84.7%. Therefore, the efficiency has not decreased.

Why is the calculation of output power so different in the solution without electrolytic capacitors (using ceramic capacitors)? The reason is that the capacitance of ceramic capacitors is small, resulting in huge ripple of output current, and the lowest value of the current has even bottomed out to zero. At this time, the ripple of the output current is already greater than its DC average value, which means that the output current is already an AC current. It is not appropriate to use the average current to calculate the output power.

The correct way to calculate the output power is: Po = Vo_rms * Io_rms * PF. In the formula, Vo_rms and Io_rms are the root mean square values ​​of the output voltage and current respectively, and PF is the power factor. Figure 4 shows the waveform and root mean square value of the output voltage and current when the output is a ceramic capacitor. Compared with Figure 3, it can be found that for AC current, the average value and the root mean square value are no longer equal.

Figure 4. Current and voltage waveforms when the output uses ceramic capacitors (22uF X 2).

(Ch1=blue: output voltage; Ch4=green: output current; math operation=red: Ch1*Ch4)

However, the power factor PF is not easy to measure. It is difficult to use the above formula. It is easier to calculate the output power by using the average value of instantaneous power (instantaneous voltage multiplied by instantaneous current). This operation can be easily achieved on an oscilloscope. In the solution using electrolytic capacitors, since the capacitance of electrolytic capacitors is relatively large, the DC value of the output current is much larger than the ripple value, and its average value is basically equal to the root mean square value. Using the average current to calculate the output power will not introduce too much error.